Problem 87 - Chemistry The Molecular Science

Question

You want to heat the air in your house with natural gas, \(\mathrm{CH}_{4}\). Assume your house has \(275 \mathrm{~m}^{2}\) (about \(2800 \mathrm{ft}^{2}\) ) of floor area and that the ceilings are \(2.50 \mathrm{~m}\) (about \(8 \mathrm{ft}\) ) from the floors. The air in the house has a molar heat capacity of \(29.1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\). (The number of moles of air in the house can be found by assuming that the average molar mass of air is \(28.9 \mathrm{~g} / \mathrm{mol}\) and that the density of air at these temperatures is \(1.22 \mathrm{~g} / \mathrm{L}\). ) Calculate what mass of methane you have to burn to heat the air from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

Approximately 118.4 g of methane is needed.

Step 1: Calculate the Volume of the House

The volume of the house can be calculated using the formula for volume, which is the floor area multiplied by the height of the ceiling. The area is given as \(275 \, \mathrm{m}^2\) and the height is \(2.50 \, \mathrm{m}\), so the volume \(V\) is:\[V = 275 \, \mathrm{m}^2 \times 2.50 \, \mathrm{m} = 687.5 \, \mathrm{m}^3\]

Step 2: Convert Volume to Liters

Since the density of air is given in grams per liter, convert the volume from cubic meters to liters. Remember that \(1 \, \mathrm{m}^3 = 1000 \, \mathrm{L}\), so:\[V = 687.5 \, \mathrm{m}^3 \times 1000 \, \mathrm{L/m}^3 = 687,500 \, \mathrm{L}\]

Step 3: Calculate the Mass of Air

Use the density of air (\(1.22 \, \mathrm{g/L}\)) to find the mass of air. Mass \(m\) is density \(\rho\) times volume \(V\):\[m = 1.22 \, \mathrm{g/L} \times 687,500 \, \mathrm{L} = 838,750 \, \mathrm{g}\]

Step 4: Calculate Moles of Air

Find the number of moles of air using its mass and molar mass \(28.9 \, \mathrm{g/mol}\):\[n = \frac{838,750 \, \mathrm{g}}{28.9 \, \mathrm{g/mol}} \approx 29027 \, \mathrm{mol}\]

Step 5: Calculate the Heat Required

The heat required \(q\) to change the temperature is given by \(q = nC_p\Delta T\), where \(C_p\) is the molar heat capacity (\(29.1 \, \mathrm{J/mol \cdot K}\)) and \(\Delta T = 22.0^{\circ} \mathrm{C} - 15.0^{\circ} \mathrm{C} = 7.0 \, \mathrm{K}\):\[q = 29027 \, \mathrm{mol} \times 29.1 \, \mathrm{J/mol \cdot K} \times 7.0 \, \mathrm{K} \approx 5,921,293 \, \mathrm{J}\]

Step 6: Determine the Energy from Combustion

The combustion of methane \(\mathrm{CH}_4\) releases roughly \(802 \, \mathrm{kJ/mol}\) of energy. Calculate how many moles of methane are needed:\[n_{\mathrm{CH}_4} = \frac{5,921,293 \, \mathrm{J}}{802,000 \, \mathrm{J/mol}} \approx 7.38 \, \mathrm{mol}\]

Step 7: Calculate Mass of Methane Needed

Find the mass of methane required using its molar mass \(16.04 \, \mathrm{g/mol}\):\[m_{\mathrm{CH}_4} = 7.38 \, \mathrm{mol} \times 16.04 \, \mathrm{g/mol} \approx 118.39 \, \mathrm{g}\]

Step 8: Conclusion

You need approximately \(118.4 \, \mathrm{g}\) of methane to heat the air in your house from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C}\).

Key Concepts

Heat Capacity

Understanding heat capacity is crucial in thermochemistry, as it denotes the amount of heat required to change the temperature of a substance by one degree Celsius. In our scenario with the house, the molar heat capacity of air is given as 29.1 J/mol·K. This value helps determine how much energy is needed to increase the temperature of the air inside your home.

The molar heat capacity reflects the energy change for air at a molecular level, affecting how much total energy must be introduced. It plays a critical role alongside the mole count of air to compute the exact amount of energy needed, ensuring accuracy in determining how efficiently the heating process via methane combustion will proceed.

To sum it up, the molar heat capacity of air is crucial for understanding how heat energy interacts with and affects the temperature of the air within a particular volume.

Molar Mass

Molar mass is a concept that bridges chemistry and physics, providing a unit mass of a substance's atoms in a mole. In the example of heating the air, the molar mass of air is averaged to 28.9 g/mol. Understanding this is fundamental in calculating how many moles of the atmospheric air are present in the home's volume.

Using molar mass, we can convert the weight of air (calculated from volume and density) into moles. This conversion is key because energy calculations are based on the number of moles and the molar heat capacity, helping to forecast the energy expenditure required to raise the temperature.

• The molar mass of air allows for bridging the gap between mass and moles.
• Accurate mass-to-mole conversions permit precise thermal energy calculations.

This conversion from macro (grams) to micro (moles) undermines the importance of molar mass in chemical thermodynamics.

Chemical Energy

Chemical energy refers to the energy stored in the bonds of chemical compounds, such as methane (CH extsubscript{4} in this case. It is from these bonds that energy is released during chemical reactions, like combustion. Understanding how chemical energy translates to usable thermal energy is pivotal when analyzing heating systems.

In our situation, the combustion of methane releases 802 kJ/mol of energy. This release of energy is critical because it determines how much actual heating is accomplished per mole of methane combusted. The process involves calculating the energy required to heat the home’s air and then juxtaposing it against this energy release from combustion. Thus, chemical energy is both the source of heat and a yardstick for fuel efficiency in this context.

Methane Combustion

Methane combustion is the chemical process of burning methane (CH extsubscript{4}) in oxygen to produce carbon dioxide, water, and heat. It is an exothermic reaction meaning it releases energy, primarily used for heating purposes. Methane's high energy content per unit makes it an effective choice for residential heating.

The formula representing this combustion is given as: \[ ext{CH}_{4} + 2 ext{O}_{2} ightarrow ext{CO}_{2} + 2 ext{H}_{2} ext{O} + ext{Heat} \]
In our exercise, around 7.38 moles of methane produced the necessary heat to elevate the home's air temperature. This ties directly into both the energy efficiency of methane combustion and the calculation's accuracy in determining the right fuel amount. Methane combustion not only keeps homes warm but also exemplifies a practical application of chemical thermodynamics, showcasing the transition from fuel to heat.